ISRO CSE 2011 | Question: 60

Question

A total of 9 units of a resource type available, and given the safe state shown below, which of the following sequence will be a safe state?

Process	Used	Max
$P_1$	2	7
$P_2$	1	6
$P_3$	2	5
$P_4$	1	4

• A. $\langle P_4, P_1, P_3, P_2\rangle$
• B. $\langle P_4, P_2, P_1, P_3\rangle $
• C. $\langle P_4, P_2, P_3, P_1\rangle $
• D. $\langle P_3, P_1, P_2, P_4 \rangle$

Answer 1

Process	Used	Max	Need
P1	2	7	5
P2	1	6	5
P3	2	5	3
P4	1	4	3

We have total 9 resources and out of 9, 6 are already allocated to the processes. Now we are left with 3 resource which can fulfill the need of process P3 and P4. Now if we check options A, B and C where first P4 is being served then after completion of P4 we will have total 3 + 1 = 4 resources but now we can't fulfill the need of either P1 or P2, because each needs 5 resource so option A, B and C can not be correct ans. Similarly we can check for option D where after P3 we can fulfill any combination. So Ans- D. <P3,P1,P2,P4><P3,P1,P2,P4><P3,P1,P2,P4>

Comments

@vijaycs  other than given options is Sequence <P4,P3,P2,P1> could be the also??

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That is a really good solution.

Thanks.

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The correct sequence (P3, P1, P2, P4) is safe because starting with P3 frees up the most resources early on. This allows enough resources to fulfill P1, followed by P2, and finally P4, ensuring no deadlock occurs.

Answer 2

According to me answer should be

⟨P3,P1,P2,P4⟩

Available resourses A=3

By this we can only fulfill the future need of the P3

Now new availbility Anew=3+2=5,

By Anew we can fill the need of P1 or P2,Suppose we fill need of P1,then Anew=5+2=7

Anew=7,Now for P2;

Anew=7+1=8

Finally with P4;

Anew=8+1=9

So our system is safe and safe sequence is P3---->P1------->P2------>P4

Comments

Perfect